洛谷 P1238 走迷宫【搜索】【DFS】

一、题目链接二、题目分析

（一）算法标签（二）解题思路 三、AC代码四、其它题解

洛谷 P1238 走迷宫

搜索 DFS

解法一:(DFS)

#include 
#include 

using namespace std;

const int N = 15;

int n, m;
int g[N][N];
bool st[N][N], has_solution;
int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};

struct Point {
	int x, y;
};

vector track;	// 存放路径 

Point start, dest;

istream& operator>> (istream &in, Point &p)
{
	in >> p.x >> p.y;
	return in;
}

ostream& operator<< (ostream &out, Point &p)
{
	out << '(' << p.x << ',' << p.y << ')';
	return out;
}

bool operator== (Point &p1, Point &p2)
{
	return p1.x == p2.x && p1.y == p2.y;
}

void printTrack()
{
	cout << track[0];
	for (int i = 1; i < track.size(); i ++ )
		cout << "->"<< track[i];
	cout << endl;
}

void dfs(Point p)
{
	if (p == dest)
	{
		has_solution = true;
		printTrack();
		return;
	}
	
	for (int i = 0; i < 4; i ++ )
	{
		int a = p.x + dx[i], b = p.y + dy[i];
		if (a < 1 || a > n || b < 1 || b > m) continue;
		if (g[a][b] == 1 && !st[a][b])
		{
			st[a][b] = true;
			track.push_back({a, b});
			dfs({a, b});
			track.pop_back();
			st[a][b] = false;
		}
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= m; j ++ )
			cin >> g[i][j];
	cin >> start >> dest;
	
	// 起点被访问，并放进track 
	st[start.x][start.y] = true;
	track.push_back(start);
	
	dfs(start);
	
	if (!has_solution) cout << "-1" << endl;
	return 0;
}

DFS另一种写法

#include 

using namespace std;

const int N = 15;

int g[N][N];
bool st[N][N];
int n, m, sx, sy, fx, fy;
int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};
bool has_solution;

void dfs(int x, int y, string s)
{
	if (x == fx && y == fy)
	{
		has_solution = true;
		cout << s << endl;
		return;
	}
	
	for (int i = 0; i < 4; i ++ )
	{
		int a = x + dx[i], b = y + dy[i];
		if (a >= 1 && a <= n && b >= 1 && b <= m 
			&& !st[a][b] && g[a][b] == 1)
		{
			st[a][b] = true;
			dfs(a, b, s + "->(" + to_string(a) + "," + to_string(b) + ")");
			st[a][b] = false;	
		}
	}
}
int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= m; j ++ )
			cin >> g[i][j];
	cin >> sx >> sy >> fx >> fy;
	st[sx][sy] = true;
	dfs(sx, sy, "(" + to_string(sx) + "," + to_string(sy) + ")");
	
	if (!has_solution) cout << "-1" << endl;
	return 0;
}

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